Use calculus to find the point on the graph where the tangent line is horizontal, for the function
the derivative of y= -0.1x^2+3x+5 is y'= -0.2x+3
set the derivative =0, so y'= -0.2x+3=0
-0.2x=-3
x=15
y= -0.1x^2+3x+5 -0.1(15)^2+3(15)+5= -22.5+45+5=27.5
the point is (15 , 27.5)
The slope of the line tangent at x is the derivative at x. A horizontal line has a slope of zero.
So, just differentiate the function and set it equal to zero and solve for x.
dy/dx = 2*(-0.1)x + 3 = 0
x = 3/0.2 = 15.
This is a downward opening parabola with it's max at x=15.
Assume equations is:
y = -0.1 * x^2 + 3*x + 5
Find slope = 0
Take derivitive:
y' = -0.2 * x + 3 = 0
x = 15
y = -0.1 * 15^2 + 3*15 + 5 = 27.5
A. 2x^3 y^2
if you meant -.1x^2+3x+5
then the horizontal line is at -b/2a because the function is parabolic
vertex = -3/-.2
vertex is at (15,27)
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the derivative of y= -0.1x^2+3x+5 is y'= -0.2x+3
set the derivative =0, so y'= -0.2x+3=0
-0.2x=-3
x=15
y= -0.1x^2+3x+5 -0.1(15)^2+3(15)+5= -22.5+45+5=27.5
the point is (15 , 27.5)
The slope of the line tangent at x is the derivative at x. A horizontal line has a slope of zero.
So, just differentiate the function and set it equal to zero and solve for x.
dy/dx = 2*(-0.1)x + 3 = 0
x = 3/0.2 = 15.
This is a downward opening parabola with it's max at x=15.
Assume equations is:
y = -0.1 * x^2 + 3*x + 5
Find slope = 0
Take derivitive:
y' = -0.2 * x + 3 = 0
x = 15
y = -0.1 * 15^2 + 3*15 + 5 = 27.5
A. 2x^3 y^2
if you meant -.1x^2+3x+5
then the horizontal line is at -b/2a because the function is parabolic
vertex = -3/-.2
vertex is at (15,27)